I got the greedy intuition pretty fast: always pick the highest-profit available project.
Start by restating the problem and clarifying constraints, then propose a greedy strategy using a min-heap to always select the most profitable project among those currently affordable. Walk through the algorithm step-by-step, analyze time and space complexity, and justify correctness with an exchange argument. Finally, discuss how the approach scales to N=200,000 and mention potential optimizations.
Pro tip: Emphasize that the greedy choice is safe because profits are positive and selecting the highest profit affordable project never reduces future affordability. Also, mention that sorting projects by capital requirement and using a min-heap for profits is a classic pattern for this type of problem.
Confirm the inputs (initial capital W, max projects K, arrays of capital requirements and profits) and the goal (maximize final capital). Clarify that you can complete at most K projects, not exactly K, and that projects can be done in any order as long as capital requirements are met.
Sort projects by capital requirement. Use a min-heap to store profits of projects that are currently affordable. At each step, add all projects with requirement ≤ current capital to the heap, then if heap is non-empty and we haven't done K projects, pop the max profit (simulate by using a max-heap or negating values) and add it to capital.
Time complexity: O(N log N) due to sorting and heap operations. Space complexity: O(N) for the heap and sorted list. Correctness: prove by exchange argument that selecting the highest profit affordable project at each step leads to an optimal solution.
Explain that O(N log N) is efficient for N=200,000. Mention edge cases: no affordable projects, K=0, all projects affordable initially, and projects with zero profit. Also, note that if K is large, the algorithm naturally stops when no more projects can be done.
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